How do you find a unit vector a)parallel to and b)normal to the graph of $f (x)$ $f(x)$ at the indicated point. $f (x) = x^{3}, at (1, 1)$ $f(x)= x^3, "at" (1,1)$ ?

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How do you find a unit vector a)parallel to and b)normal to the graph of $f (x)$ $f(x)$ at the indicated point. $f (x) = x^{3}, at (1, 1)$ $f(x)= x^3, "at" (1,1)$ ?

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Answer 1 · 2017-10-03T08:41:47

Unit vector patallel to function at $(1, 1)$ $(1,1)$ is $\frac{3}{\sqrt{10}} ˆ i - \frac{1}{\sqrt{10}} ˆ j$ $3/sqrt10hati-1/sqrt10hatj$ and normal vector is $\frac{1}{\sqrt{10}} ˆ i + \frac{3}{\sqrt{10}} ˆ j$ $1/sqrt10hati+3/sqrt10hatj$

Explanation:

The slope of tangent at $f (x) = x^{3}$ $f(x)=x^3$ at $(1, 1)$ $(1,1)$ is $f'(1)$ and hence slope of normal is $-1/(f'(1))$

Now as $f(x)=x^3$ , $f'(x)=3x^2$ . Slope of tangent at $(1,1)$ is $3$

and slope of normal is $-1/3$

Therefore equation of tangent is $y-1=3(x-1)$ or $3x-y=2$

and unit vector parallel to $f(x)$ would be $(3hati-hatj)/sqrt(1^+3^2)$

or $3/sqrt10hati-1/sqrt10hatj$

and equation of normal is $y-1=-1/3(x-1)$

i.e. $3y-3=-x+1$ or $x+3y=4$

and unit normal vector would be $(hati+3hatj)/sqrt(1^+3^2)$

or $1/sqrt10hati+3/sqrt10hatj$ , where $hati$ and $hatj$ are unit vectors along $x$ -axis and $y$ -axis respectively.

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How do you find a unit vector a)parallel to and b)normal to the graph of $f (x)$ $f(x)$ at the indicated point. $f (x) = x^{3}, at (1, 1)$ $f(x)= x^3, "at" (1,1)$ ?

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Explanation:

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How do you find a unit vector a)parallel to and b)normal to the graph of f(x)f(x)f(x) at the indicated point. f(x)= x^3, "at" (1,1)f(x)=x3,at(1,1)f(x)= x^3, "at" (1,1)?

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Explanation:

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How do you find a unit vector a)parallel to and b)normal to the graph of $f (x)$ $f(x)$ at the indicated point. $f (x) = x^{3}, at (1, 1)$ $f(x)= x^3, "at" (1,1)$ ?